# Symplectic form for real solutions to the Klein-Gordon equation in curved spacetime

Consider a real Klein-Gordon field $$\phi$$ in a globally hyperbolic spacetime, with metric $$g_{\mu\nu}$$.

The covariant Klein-Gordon equation is

$$(g^{\mu\nu}\nabla_{\mu}\nabla_{\nu}+m^{2})\phi=0$$

Let $$V$$ be the vector space of all real-valued solutions to this equation.

Define the following bilinear form:

$$(f,h):=\int_{\Sigma} d\Sigma_{\mu}\sqrt g g^{\mu\nu} (f\partial_{\nu}h-h\partial_{\nu}f), \forall f,h \in V$$

where $$\Sigma$$ is any spacelike hypersurface.

This is a bilinear, antisymmetric, non-degenerate form, i.e. a symplectic form.

How can I prove that it is non-degenerate?

• Probably some kind of bump function. If f is nonzero somewhere, try picking some h which is 1 nearby and 0 elsewhere. Probably have to be a bit more careful since there are derivatives involved, try integrating the first bit by parts maybe? Commented Jan 4 at 16:16
• I thought maybe you can find a set of basis $\{f_1,...,f_n\}$, and rewrite $f$ and $h$ by $f=a_1 f_1+...+a_n f_n$, $h=b_1 f_1+...+b_n f_n$ (it could be non-discrete), and then show that only when $a_1=a_2=...=a_n=0$ or $b_1=b_2=...=b_n=0$, the inner product could be zero. Commented Jan 5 at 4:12
• In this particular case, if $f=0$ whenever $(f,h) = 0$ for all of $h$, then the inner product you wrote down is nondegenerate. This is the usual definition, so I am more than likely missing some details since it is bilinear (like something about isomorphisms between vector spaces) but this is the bare bone version. Commented Jan 5 at 16:15

It is easy actually. To be precise $$\Sigma$$ is a smooth spacelike Cauchy surface of the spacetime and the considered space $$V$$ of solutions of the KG equations is made of solutions smooth and with compactly supported Cauchy data on $$\Sigma$$. Under these hypotheses, the relation between Cauchy data on $$\Sigma$$, $$(f|_\Sigma, n_\Sigma \cdot\nabla f|_\Sigma)$$ and corresponding solutions $$f$$ of the KG equation is one-to-one. In other words the Cauchy problem is well posed.
If $$(f,h)=0$$ for every $$h$$ then both $$f$$ and its derivative normal to $$\Sigma$$ are zero. This easily follows from the very form of the simplectic form and from the fact that we can choose the Cauchy data of $$h$$ arbitrarily and there is a corresponding $$h$$.
Hence, again in view of the well posedness of the Cauchy problem, $$f$$ must be the zero solution of the KG equation.
• $(f,h)=0, \forall h$ implies, for the well posedness, that the integral is $0$ for every possible pair $(h|_\Sigma, n_\Sigma \cdot\nabla h|_\Sigma)$, which implies that $(f|_\Sigma, n_\Sigma \cdot\nabla f|_\Sigma)=(0,0)$, which implies, for the well posedness, that $f=0$. Right?